{"id":1954,"date":"2023-10-12T00:36:10","date_gmt":"2023-10-12T00:36:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/vectors\/"},"modified":"2024-11-20T23:05:44","modified_gmt":"2024-11-20T23:05:44","slug":"vectors","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/vectors\/","title":{"raw":"Vectors","rendered":"Vectors"},"content":{"raw":"<div id=\"proctor-360-confirm\" class=\"proctor-360-confirm proctor-hidden\" style=\"display: none;\">\r\n\r\n<img class=\"proctor-360-login-img text-center\" style=\"width: 50px;\" src=\"chrome-extension:\/\/hkegehhbmbongohpgmdadkbkmnfokicn\/img\/icon128.png\" alt=\"logo\" \/>\r\n<h3 class=\"text-center\">Are you sure to stop exam tracking?<\/h3>\r\n<div class=\"proctor-actions\"><button id=\"proctor-360-stop-exam\" class=\"proctor-360-stop-exam btn btn-danger\" type=\"submit\">Yes<\/button> <button id=\"proctor-360-cancel\" class=\"proctor-360-cancel btn btn-success\" type=\"submit\">No<\/button><\/div>\r\n<\/div>\r\n<div id=\"proctor-cntnr\"><\/div>\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>View vectors geometrically.<\/li>\r\n \t<li>Find magnitude and direction.<\/li>\r\n \t<li>Perform vector addition and scalar multiplication.<\/li>\r\n \t<li>Find the component form of a vector.<\/li>\r\n \t<li>Find the unit vector in the direction of [latex]\\boldsymbol{v}[\/latex] .<\/li>\r\n \t<li>Perform operations with vectors in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex] .<\/li>\r\n \t<li>Find the dot product of two vectors.<\/li>\r\n<\/ul>\r\n<\/div>\r\nAn airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140\u00b0. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 1. What are the ground speed and actual bearing of the plane?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181127\/CNX_Precalc_Figure_08_08_0012.jpg\" alt=\"Image of a plan flying SE at 140 degrees and the north wind blowing \" width=\"487\" height=\"462\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\nGround speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane\u2019s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let\u2019s examine the basics of vectors.\r\n<h2>A Geometric View of Vectors<\/h2>\r\nA <strong>vector<\/strong> is a specific quantity drawn as a line segment with an arrowhead at one end. It has an <strong>initial point<\/strong>, where it begins, and a <strong>terminal point<\/strong>, where it ends. A vector is defined by its <strong>magnitude<\/strong>, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:\r\n<ul>\r\n \t<li>Lower case, boldfaced type, with or without an arrow on top such as [latex]\\boldsymbol{v,u,w,\\stackrel{\\to }{v},\\stackrel{\\to }{u},\\stackrel{\\to }{w}}[\/latex].<\/li>\r\n \t<li>Given initial point [latex]P[\/latex] and terminal point [latex]Q[\/latex], a vector can be represented as [latex]\\stackrel{\\to }{PQ}[\/latex]. The arrowhead on top is what indicates that it is not just a line, but a directed line segment.<\/li>\r\n \t<li>Given an initial point of [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(a,b\\right)[\/latex], a vector may be represented as [latex]\\langle a,b\\rangle [\/latex].<\/li>\r\n<\/ul>\r\nThis last symbol [latex]\\langle a,b\\rangle [\/latex] has special significance. It is called the <strong>standard position<\/strong>. The <strong>position vector<\/strong> has an initial point [latex]\\left(0,0\\right)[\/latex] and a terminal point [latex]\\langle a,b\\rangle [\/latex]. To change any vector into the position vector, we think about the change in the <em>x<\/em>-coordinates and the change in the <em>y<\/em>-coordinates. Thus, if the initial point of a vector [latex]\\stackrel{\\to }{CD}[\/latex] is [latex]C\\left({x}_{1},{y}_{1}\\right)[\/latex] and the terminal point is [latex]D\\left({x}_{2},{y}_{2}\\right)[\/latex], then the position vector is found by calculating\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\stackrel{\\to }{AB}&amp;=\\langle {x}_{2}-{x}_{1},{y}_{2}-{y}_{1}\\rangle \\\\ &amp;=\\langle a,b\\rangle \\end{align}[\/latex]<\/div>\r\nIn Figure 2, we see the original vector [latex]\\stackrel{\\to }{CD}[\/latex] and the position vector [latex]\\stackrel{\\to }{AB}[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181129\/CNX_Precalc_Figure_08_08_0032.jpg\" alt=\"Plot of the original vector CD in blue and the position vector AB in orange extending from the origin.\" width=\"487\" height=\"290\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Properties of Vectors<\/h3>\r\nA vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at [latex]\\left(0,0\\right)[\/latex] and is identified by its terminal point [latex]\\langle a,b\\rangle [\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Find the Position Vector<\/h3>\r\nConsider the vector whose initial point is [latex]P\\left(2,3\\right)[\/latex] and terminal point is [latex]Q\\left(6,4\\right)[\/latex]. Find the position vector.\r\n\r\n[reveal-answer q=\"454005\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"454005\"]\r\n\r\nThe position vector is found by subtracting one <em>x<\/em>-coordinate from the other <em>x<\/em>-coordinate, and one <em>y<\/em>-coordinate from the other <em>y<\/em>-coordinate. Thus\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&amp;=\\langle 6 - 2,4 - 3\\rangle \\\\ &amp;=\\langle 4,1\\rangle \\end{align}[\/latex]<\/p>\r\nThe position vector begins at [latex]\\left(0,0\\right)[\/latex] and terminates at [latex]\\left(4,1\\right)[\/latex]. The graphs of both vectors are shown in Figure 3.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181132\/CNX_Precalc_Figure_08_08_0222.jpg\" alt=\"Plot of the original vector in blue and the position vector in orange extending from the origin.\" width=\"487\" height=\"349\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\nWe see that the position vector is [latex]\\langle 4,1\\rangle [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Drawing a Vector with the Given Criteria and Its Equivalent Position Vector<\/h3>\r\nFind the position vector given that vector<em><strong> v <\/strong><\/em>has an initial point at [latex]\\left(-3,2\\right)[\/latex] and a terminal point at [latex]\\left(4,5\\right)[\/latex], then graph both vectors in the same plane.\r\n\r\n[reveal-answer q=\"823850\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"823850\"]\r\n\r\nThe position vector is found using the following calculation:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&amp;=\\langle 4-\\left(-3\\right),5 - 2\\rangle \\\\ &amp;=\\langle 7,3\\rangle \\end{align}[\/latex]<\/p>\r\nThus, the position vector begins at [latex]\\left(0,0\\right)[\/latex] and terminates at [latex]\\left(7,3\\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181134\/CNX_Precalc_Figure_08_08_004n2.jpg\" alt=\"Plot of the two given vectors their same position vector.\" width=\"487\" height=\"328\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nDraw a vector [latex]\\boldsymbol{v}[\/latex] that connects from the origin to the point [latex]\\left(3,5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"884783\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"884783\"]\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181235\/CNX_Precalc_Figure_08_08_0062.jpg\" alt=\"A vector from the origin to (3,5) - a line with an arrow at the (3,5) endpoint.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173913[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Finding Magnitude and Direction<\/h2>\r\nTo work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Magnitude and Direction of a Vector<\/h3>\r\nGiven a position vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle [\/latex], the magnitude is found by [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]. The direction is equal to the angle formed with the <em>x<\/em>-axis, or with the <em>y<\/em>-axis, depending on the application. For a position vector, the direction is found by [latex]\\tan \\theta =\\left(\\frac{b}{a}\\right)\\Rightarrow \\theta ={\\tan }^{-1}\\left(\\frac{b}{a}\\right)[\/latex], as illustrated in Figure 5.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181137\/CNX_Precalc_Figure_08_08_017new2.jpg\" alt=\"Standard plot of a position vector (a,b) with magnitude |v| extending into Q1 at theta degrees. \" width=\"487\" height=\"216\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\nTwo vectors <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong> are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Finding the Magnitude and Direction of a Vector<\/h3>\r\nFind the magnitude and direction of the vector with initial point [latex]P\\left(-8,1\\right)[\/latex] and terminal point [latex]Q\\left(-2,-5\\right)[\/latex]. Draw the vector.\r\n\r\n[reveal-answer q=\"241600\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"241600\"]\r\n\r\nFirst, find the <strong>position vector<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{u}&amp;=\\langle -2,-\\left(-8\\right),-5 - 1\\rangle \\\\ &amp;=\\langle 6,-6\\rangle \\end{align}[\/latex]<\/p>\r\nWe use the Pythagorean Theorem to find the magnitude.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{u}|&amp;=\\sqrt{{\\left(6\\right)}^{2}+{\\left(-6\\right)}^{2}} \\\\ &amp;=\\sqrt{72} \\\\ &amp;=6\\sqrt{2} \\end{align}[\/latex]<\/p>\r\nThe direction is given as\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\tan \\theta =\\frac{-6}{6}=-1\\\\ &amp;\\theta ={\\tan }^{-1}\\left(-1\\right) =-45^\\circ \\end{align}[\/latex]<\/p>\r\nHowever, the angle terminates in the fourth quadrant, so we add 360\u00b0 to obtain a positive angle. Thus, [latex]-45^\\circ +360^\\circ =315^\\circ [\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181140\/CNX_Precalc_Figure_08_08_0182.jpg\" alt=\"Plot of the position vector extending into Q4 from the origin with the magnitude 6rad2.\" width=\"487\" height=\"316\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Showing That Two Vectors Are Equal<\/h3>\r\nShow that vector <strong><em>v<\/em><\/strong> with <strong>initial point<\/strong> at [latex]\\left(5,-3\\right)[\/latex] and <strong>terminal point<\/strong> at [latex]\\left(-1,2\\right)[\/latex] is equal to vector <strong><em>u<\/em><\/strong> with initial point at [latex]\\left(-1,-3\\right)[\/latex] and terminal point at [latex]\\left(-7,2\\right)[\/latex]. Draw the position vector on the same grid as <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong>. Next, find the magnitude and direction of each vector.\r\n\r\n[reveal-answer q=\"46978\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"46978\"]\r\n\r\nAs shown in Figure 7, draw the vector [latex]\\boldsymbol{v}[\/latex] starting at initial [latex]\\left(5,-3\\right)[\/latex] and terminal point [latex]\\left(-1,2\\right)[\/latex]. Draw the vector [latex]\\boldsymbol{u}[\/latex] with initial point [latex]\\left(-1,-3\\right)[\/latex] and terminal point [latex]\\left(-7,2\\right)[\/latex]. Find the standard position for each.\r\n\r\nNext, find and sketch the position vector for <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong>. We have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&amp;=\\langle -1 - 5,2-\\left(-3\\right)\\rangle \\\\ &amp;=\\langle -6,5\\rangle \\\\ \\text{ } \\\\ \\boldsymbol{u}&amp;=\\langle -7-\\left(-1\\right),2-\\left(-3\\right)\\rangle \\\\ &amp;=\\langle -6,5\\rangle \\end{align}[\/latex]<\/p>\r\nSince the position vectors are the same, <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong> are the same.\r\n\r\nAn alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{v}|&amp;=\\sqrt{{\\left(-1 - 5\\right)}^{2}+{\\left(2-\\left(-3\\right)\\right)}^{2}} \\\\ &amp;=\\sqrt{{\\left(-6\\right)}^{2}+{\\left(5\\right)}^{2}} \\\\ &amp;=\\sqrt{36+25} \\\\ &amp;=\\sqrt{61} \\\\ |\\boldsymbol{u}|&amp;=\\sqrt{{\\left(-7-\\left(-1\\right)\\right)}^{2}+{\\left(2-\\left(-3\\right)\\right)}^{2}} \\\\ &amp;=\\sqrt{{\\left(-6\\right)}^{2}+{\\left(5\\right)}^{2}} \\\\ &amp;=\\sqrt{36+25} \\\\ &amp;=\\sqrt{61} \\end{align}[\/latex]<\/p>\r\nAs the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\tan \\theta =-\\frac{5}{6}\\\\ &amp;\\theta ={\\tan }^{-1}\\left(-\\frac{5}{6}\\right) =-39.8^\\circ \\end{align}[\/latex]<\/p>\r\nHowever, we can see that the position vector terminates in the second quadrant, so we add [latex]180^\\circ [\/latex]. Thus, the direction is [latex]-39.8^\\circ +180^\\circ =140.2^\\circ [\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181143\/CNX_Precalc_Figure_08_08_005n2.jpg\" alt=\"Plot of the two given vectors their same position vector.\" width=\"487\" height=\"440\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Performing Vector Addition and Scalar Multiplication<\/h2>\r\nNow that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector [latex]\\boldsymbol{u}[\/latex] [latex]=\\langle x,y\\rangle [\/latex] as an arrow or directed line segment from the origin to the point [latex]\\left(x,y\\right)[\/latex], vectors can be situated anywhere in the plane. The sum of two vectors <strong><em>u<\/em><\/strong> and <strong><em>v<\/em><\/strong>, or <strong>vector addition<\/strong>, produces a third vector <strong><em>u<\/em><\/strong>+ <strong><em>v<\/em><\/strong>, the <strong>resultant<\/strong> vector.\r\n\r\nTo find <strong><em>u<\/em><\/strong> + <strong><em>v<\/em><\/strong>, we first draw the vector <strong><em>u<\/em><\/strong>, and from the terminal end of <strong><em>u<\/em><\/strong>, we draw the vector <strong><em>v<\/em><\/strong>. In other words, we have the initial point of <strong><em>v<\/em><\/strong> meet the terminal end of <strong><em>u<\/em><\/strong>. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum <strong><em>u<\/em><\/strong> + <strong><em>v<\/em><\/strong> is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of <strong><em>u<\/em><\/strong> to the end of <strong><em>v<\/em><\/strong> in a straight path, as shown in Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181145\/CNX_Precalc_Figure_08_08_0082.jpg\" alt=\"Diagrams of vector addition and subtraction. \" width=\"487\" height=\"149\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\nVector subtraction is similar to vector addition. To find <strong><em>u<\/em><\/strong> \u2212 <strong><em>v<\/em><\/strong>, view it as <strong><em>u<\/em><\/strong> + (\u2212<strong><em>v<\/em><\/strong>). Adding \u2212<strong><em>v<\/em><\/strong> is reversing direction of <strong><em>v<\/em><\/strong> and adding it to the end of <strong><em>u<\/em><\/strong>. The new vector begins at the start of <strong><em>u<\/em><\/strong> and stops at the end point of \u2212<strong><em>v<\/em><\/strong>. See Figure 9\u00a0for a visual that compares vector addition and vector subtraction using <strong>parallelograms<\/strong>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181147\/CNX_Precalc_Figure_08_08_0092.jpg\" alt=\"Showing vector addition and subtraction with parallelograms. For addition, the base is u, the side is v, the diagonal connecting the start of the base to the end of the side is u+v. For subtraction, thetop is u, the side is -v, and the diagonal connecting the start of the top to the end of the side is u-v.\" width=\"487\" height=\"128\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Adding and Subtracting Vectors<\/h3>\r\nGiven [latex]\\boldsymbol{u}[\/latex] [latex]=\\langle 3,-2\\rangle [\/latex] and [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle -1,4\\rangle [\/latex], find two new vectors <strong><em>u<\/em><\/strong> + <strong><em>v<\/em><\/strong>, and <strong><em>u<\/em><\/strong> \u2212 <strong>v<\/strong>.\r\n\r\n[reveal-answer q=\"664138\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"664138\"]\r\n\r\nTo find the sum of two vectors, we add the components. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{u}+\\boldsymbol{v}&amp;=\\langle 3,-2\\rangle +\\langle -1,4\\rangle \\\\ &amp;=\\langle 3+\\left(-1\\right),-2+4\\rangle \\\\ &amp;=\\langle 2,2\\rangle \\end{align}[\/latex]<\/p>\r\nSee Figure 10(a).\r\n\r\nTo find the difference of two vectors, add the negative components of [latex]v[\/latex] to [latex]u[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{u}+\\left(-\\boldsymbol{v}\\right)&amp;=\\langle 3,-2\\rangle +\\langle 1,-4\\rangle \\\\ &amp;=\\langle 3+1,-2+\\left(-4\\right)\\rangle \\\\ &amp;=\\langle 4,-6\\rangle \\end{align}[\/latex]<\/p>\r\nSee Figure 10(b).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181150\/CNX_Precalc_Figure_08_08_0192.jpg\" alt=\"Further diagrams of vector addition and subtraction.\" width=\"731\" height=\"292\" \/> <b>Figure 10.<\/b> (a) Sum of two vectors (b) Difference of two vectors[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Multiplying By a Scalar<\/h2>\r\nWhile adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a <strong>scalar<\/strong>, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Scalar Multiplication<\/h3>\r\n<strong>Scalar multiplication<\/strong> involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle [\/latex] by [latex]k[\/latex] , we have\r\n<p style=\"text-align: center;\">[latex]k\\boldsymbol{v}=\\langle ka,kb\\rangle [\/latex]<\/p>\r\nOnly the magnitude changes, unless [latex]k[\/latex] is negative, and then the vector reverses direction.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Performing Scalar Multiplication<\/h3>\r\nGiven vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle 3,1\\rangle [\/latex], find 3<strong><em>v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] <strong><em>v<\/em>, <\/strong>and \u2212<em><strong>v<\/strong><\/em>.\r\n\r\n[reveal-answer q=\"962766\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"962766\"]\r\n\r\nSee Figure 11\u00a0for a geometric interpretation. If [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle 3,1\\rangle [\/latex], then\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3\\boldsymbol{v}&amp;=\\langle 3\\cdot 3,3\\cdot 1\\rangle \\\\ &amp;=\\langle 9,3\\rangle \\\\ \\frac{1}{2}\\boldsymbol{v}&amp;=\\langle \\frac{1}{2}\\cdot 3,\\frac{1}{2}\\cdot 1\\rangle \\\\ &amp;=\\langle \\frac{3}{2},\\frac{1}{2}\\rangle \\\\ -\\boldsymbol{v}&amp;=\\langle -3,-1\\rangle \\end{align}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181152\/CNX_Precalc_Figure_08_08_0072.jpg\" alt=\"Showing the effect of scaling a vector: 3x, 1x, .5x, and -1x. The 3x is three times as long, the 1x stays the same, the .5x halves the length, and the -1x reverses the direction of the vector but keeps the length the same. The rest keep the same direction; only the magnitude changes.\" width=\"487\" height=\"367\" \/> <b>Figure 11<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice that the vector 3<strong><em>v<\/em><\/strong> is three times the length of <strong><em>v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] [latex]\\boldsymbol{v}[\/latex] is half the length of <strong><em>v<\/em><\/strong>, and \u2013<strong><em>v<\/em><\/strong> is the same length of <strong><em>v<\/em><\/strong>, but in the opposite direction.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the <strong>scalar multiple<\/strong>\u00a0[latex]3\\boldsymbol{u}[\/latex] given [latex]\\boldsymbol{u}[\/latex] [latex]=\\langle 5,4\\rangle [\/latex].\r\n\r\n[reveal-answer q=\"453643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453643\"]\r\n\r\n[latex]3\\boldsymbol{u}=\\langle 15,12\\rangle [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Using Vector Addition and Scalar Multiplication to Find a New Vector<\/h3>\r\nGiven [latex]\\boldsymbol{u}=\\langle 3,-2\\rangle [\/latex] and [latex]\\boldsymbol{v}=\\langle -1,4\\rangle [\/latex], find a new vector <strong><em>w<\/em><\/strong> = 3<strong><em>u<\/em><\/strong> + 2<strong>v<\/strong>.\r\n\r\n[reveal-answer q=\"658063\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"658063\"]\r\n\r\nFirst, we must multiply each vector by the scalar.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3\\boldsymbol{u}&amp;=3\\langle 3,-2\\rangle \\\\ &amp;=\\langle 9,-6\\rangle \\\\ 2\\boldsymbol{v}&amp;=2\\langle -1,4\\rangle \\\\ &amp;=\\langle -2,8\\rangle \\end{align}[\/latex]<\/p>\r\nThen, add the two together.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{w}&amp;=3\\boldsymbol{u}+2\\boldsymbol{v} \\\\ &amp;=\\langle 9,-6\\rangle +\\langle -2,8\\rangle \\\\ &amp;=\\langle 9 - 2,-6+8\\rangle \\\\ &amp;=\\langle 7,2\\rangle \\end{align}[\/latex]<\/p>\r\nSo, [latex]\\boldsymbol{w}=\\langle 7,2\\rangle [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173921[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding Component Form<\/h2>\r\nIn some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the [latex]x[\/latex] direction, and the vertical component is the [latex]y[\/latex] direction. For example, we can see in the graph in Figure 12\u00a0that the position vector [latex]\\langle 2,3\\rangle [\/latex] comes from adding the vectors <strong><em>v<\/em><\/strong><sub>1<\/sub> and <strong><em>v<\/em><\/strong><sub>2<\/sub>. We have <strong><em>v<\/em><\/strong><sub>1<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(2,0\\right)[\/latex].\r\n<div>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{1}&amp;=\\langle 2 - 0,0 - 0\\rangle \\\\ &amp;=\\langle 2,0\\rangle \\end{align}[\/latex]<\/p>\r\nWe also have <strong><em>v<\/em><\/strong><sub>2<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(0,3\\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{2}&amp;=\\langle 0 - 0,3 - 0\\rangle \\\\ &amp;=\\langle 0,3\\rangle \\end{align}[\/latex]<\/div>\r\nTherefore, the position vector is\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&amp;=\\langle 2+0,3+0\\rangle \\\\ &amp;=\\langle 2,3\\rangle \\end{align}[\/latex]<\/div>\r\nUsing the Pythagorean Theorem, the magnitude of <strong><em>v<\/em><\/strong><sub>1<\/sub> is 2, and the magnitude of <strong><em>v<\/em><\/strong><sub>2<\/sub> is 3. To find the magnitude of <strong><em>v<\/em><\/strong>, use the formula with the position vector.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{v}|&amp;=\\sqrt{|{\\boldsymbol{v}}_{1}{|}^{2}+|{\\boldsymbol{v}}_{2}{|}^{2}} \\\\ &amp;=\\sqrt{{2}^{2}+{3}^{2}} \\\\ &amp;=\\sqrt{13} \\end{align}[\/latex]<\/div>\r\nThe magnitude of <strong><em>v<\/em><\/strong> is [latex]\\sqrt{13}[\/latex]. To find the direction, we use the tangent function [latex]\\tan \\theta =\\frac{y}{x}[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;\\tan \\theta =\\frac{{\\boldsymbol{v}}_{2}}{{\\boldsymbol{v}}_{1}} \\\\ &amp;\\tan \\theta =\\frac{3}{2} \\\\ &amp;\\theta ={\\tan }^{-1}\\left(\\frac{3}{2}\\right)=56.3^\\circ \\end{align}[\/latex]<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181154\/CNX_Precalc_Figure_08_08_0202.jpg\" alt=\"Diagram of a vector in root position with its horizontal and vertical components.\" width=\"487\" height=\"289\" \/> <b>Figure 12<\/b>[\/caption]\r\n\r\nThus, the magnitude of [latex]\\boldsymbol{v}[\/latex] is [latex]\\sqrt{13}[\/latex] and the direction is [latex]{56.3}^{\\circ }[\/latex] off the horizontal.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Finding the Components of the Vector<\/h3>\r\nFind the components of the vector [latex]\\boldsymbol{v}[\/latex] with initial point [latex]\\left(3,2\\right)[\/latex] and terminal point [latex]\\left(7,4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"171727\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"171727\"]\r\n\r\nFirst find the standard position.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&amp;=\\langle 7 - 3,4 - 2\\rangle \\\\ &amp;=\\langle 4,2\\rangle \\end{align}[\/latex]<\/p>\r\nSee the illustration in Figure 13.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181157\/CNX_Precalc_Figure_08_08_0212.jpg\" alt=\"Diagram of a vector in root position with its horizontal (4,0) and vertical (0,2) components.\" width=\"487\" height=\"254\" \/> <b>Figure 13<\/b>[\/caption]\r\n\r\nThe horizontal component is [latex]{\\boldsymbol{v}}_{1}=\\langle 4,0\\rangle [\/latex] and the vertical component is [latex]{\\boldsymbol{v}}_{2}=\\langle 0,2\\rangle [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>\u00a0Finding the Unit Vector in the Direction of v<\/h2>\r\nIn addition to finding a vector\u2019s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a <strong>unit vector<\/strong>. We can then preserve the direction of the original vector while simplifying calculations.\r\n\r\nUnit vectors are defined in terms of components. The horizontal unit vector is written as [latex]\\boldsymbol{i}=\\langle 1,0\\rangle [\/latex] and is directed along the positive horizontal axis. The vertical unit vector is written as [latex]\\boldsymbol{j}=\\langle 0,1\\rangle [\/latex] and is directed along the positive vertical axis.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181159\/CNX_Precalc_Figure_08_08_011n2.jpg\" alt=\"Plot showing the unit vectors i=91,0) and j=(0,1)\" width=\"487\" height=\"253\" \/> <b>Figure 14<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Unit Vectors<\/h3>\r\nIf [latex]\\boldsymbol{v}[\/latex] is a nonzero vector, then [latex]\\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}[\/latex] is a unit vector in the direction of [latex]\\boldsymbol{v}[\/latex]. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Finding the Unit Vector in the Direction of <em>v<\/em><\/h3>\r\nFind a unit vector in the same direction as [latex]\\boldsymbol{v}=\\langle -5,12\\rangle [\/latex].\r\n\r\n[reveal-answer q=\"326943\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"326943\"]\r\n\r\nFirst, we will find the magnitude.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{v}|&amp;=\\sqrt{{\\left(-5\\right)}^{2}+{\\left(12\\right)}^{2}} \\\\ &amp;=\\sqrt{25+144} \\\\ &amp;=\\sqrt{169} \\\\ &amp;=13\\end{align}[\/latex]<\/p>\r\nThen we divide each component by [latex]|v|[\/latex], which gives a unit vector in the same direction as <strong>v<\/strong>:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}=-\\frac{5}{13}i+\\frac{12}{13}j[\/latex]<\/p>\r\nor, in component form\r\n<p style=\"text-align: center;\">[latex]\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}=\\langle -\\frac{5}{13},\\frac{12}{13}\\rangle [\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181202\/CNX_Precalc_Figure_08_08_0122.jpg\" alt=\"Plot showing the unit vector (-5\/13, 12\/13) in the direction of (-5, 12)\" width=\"487\" height=\"628\" \/> <b>Figure 15<\/b>[\/caption]\r\n\r\nVerify that the magnitude of the unit vector equals 1. The magnitude of [latex]-\\frac{5}{13}\\boldsymbol{i}+\\frac{12}{13}\\boldsymbol{j}[\/latex] is given as\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{{\\left(-\\frac{5}{13}\\right)}^{2}+{\\left(\\frac{12}{13}\\right)}^{2}}&amp;=\\sqrt{\\frac{25}{169}+\\frac{144}{169}} \\\\ &amp;=\\sqrt{\\frac{169}{169}}=1 \\end{align}[\/latex]<\/p>\r\nThe vector\u00a0[latex]u=\\frac{-5}{13}[\/latex] <strong><em>i<\/em><\/strong> [latex]+\\frac{12}{13}[\/latex] <strong><em>j<\/em><\/strong> is the unit vector in the same direction as <strong><em>v<\/em><\/strong> [latex]=\\langle -5,12\\rangle [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Performing Operations with Vectors in Terms of i and j<\/h2>\r\nSo far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of <strong><em>i<\/em><\/strong> and <strong>j<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Vectors in the Rectangular Plane<\/h3>\r\nGiven a vector<strong> [latex]v[\/latex] <\/strong>with initial point [latex]P=\\left({x}_{1},{y}_{1}\\right)[\/latex] and terminal point [latex]Q=\\left({x}_{2},{y}_{2}\\right)[\/latex], <strong><em>v<\/em><\/strong> is written as\r\n<p style=\"text-align: center;\">[latex]\\boldsymbol{v}=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{1}-{y}_{2}\\right)\\boldsymbol{j}[\/latex]<\/p>\r\nThe position vector from [latex]\\left(0,0\\right)[\/latex] to [latex]\\left(a,b\\right)[\/latex], where [latex]\\left({x}_{2}-{x}_{1}\\right)=a[\/latex] and [latex]\\left({y}_{2}-{y}_{1}\\right)=b[\/latex], is written as <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em>. This vector sum is called a linear combination of the vectors <strong><em>i<\/em><\/strong> and <strong>j<\/strong>.\r\n\r\nThe magnitude of <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em> is given as [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181204\/CNX_Precalc_Figure_08_08_010new2.jpg\" alt=\"Plot showing vectors in rectangular coordinates in terms of i and j. The position vector v (in orange) extends from the origin to some point (a,b) in Q1. The horizontal (ai) and vertical (bj) components are shown.\" width=\"487\" height=\"237\" \/> <b>Figure 16<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Writing a Vector in Terms of <em>i<\/em> and <em>j<\/em><\/h3>\r\nGiven a vector [latex]\\boldsymbol{v}[\/latex] with initial point [latex]P=\\left(2,-6\\right)[\/latex] and terminal point [latex]Q=\\left(-6,6\\right)[\/latex], write the vector in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex]<strong>.<\/strong>\r\n\r\n[reveal-answer q=\"997243\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"997243\"]\r\n\r\nBegin by writing the general form of the vector. Then replace the coordinates with the given values.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&amp;=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{2}-{y}_{1}\\right)\\boldsymbol{j} \\\\ &amp;=\\left(-6 - 2\\right)\\boldsymbol{i}+\\left(6-\\left(-6\\right)\\right)\\boldsymbol{j} \\\\ &amp;=-8\\boldsymbol{i}+12\\boldsymbol{j} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Writing a Vector in Terms of <em>i<\/em> and <em>j<\/em> Using Initial and Terminal Points<\/h3>\r\nGiven initial point [latex]{P}_{1}=\\left(-1,3\\right)[\/latex] and terminal point [latex]{P}_{2}=\\left(2,7\\right)[\/latex], write the vector [latex]\\boldsymbol{v}[\/latex] in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex].\r\n\r\n[reveal-answer q=\"301789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"301789\"]\r\n\r\nBegin by writing the general form of the vector. Then replace the coordinates with the given values.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&amp;=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{2}-{y}_{1}\\right)\\boldsymbol{j} \\\\ &amp;=\\left(2-\\left(-1\\right)\\right)\\boldsymbol{i}+\\left(7 - 3\\right)\\boldsymbol{j} \\\\ &amp;=3\\boldsymbol{i}+4\\boldsymbol{j} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the vector [latex]\\boldsymbol{u}[\/latex] with initial point [latex]P=\\left(-1,6\\right)[\/latex] and terminal point [latex]Q=\\left(7,-5\\right)[\/latex] in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex].\r\n\r\n[reveal-answer q=\"445026\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"445026\"]\r\n\r\n[latex]\\boldsymbol{u}=8\\boldsymbol{i} - 11\\boldsymbol{j}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Performing Operations on Vectors in Terms of <em>i<\/em> and <em>j<\/em><\/h2>\r\nWhen vectors are written in terms of<em><strong> i <\/strong><\/em>and<strong><em> j<\/em><\/strong>, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Adding and Subtracting Vectors in Rectangular Coordinates<\/h3>\r\nGiven <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em> and <strong><em>u<\/em><\/strong> = <em>c<strong>i<\/strong><\/em> + <em>d<strong>j<\/strong><\/em>, then\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\boldsymbol{v}+\\boldsymbol{u}=\\left(a+c\\right)\\boldsymbol{i}+\\left(b+d\\right)\\boldsymbol{j}\\\\ \\boldsymbol{v}-\\boldsymbol{u}=\\left(a-c\\right)\\boldsymbol{i}+\\left(b-d\\right)\\boldsymbol{j}\\end{gathered}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 12: Finding the Sum of the Vectors<\/h3>\r\nFind the sum of [latex]{\\boldsymbol{v}}_{1}=2\\boldsymbol{i} - 3\\boldsymbol{j}[\/latex] and [latex]{\\boldsymbol{v}}_{2}=4\\boldsymbol{i}+5\\boldsymbol{j}[\/latex].\r\n\r\n[reveal-answer q=\"139436\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"139436\"]\r\n\r\nAccording to the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{1}+{\\boldsymbol{v}}_{2}&amp;=\\left(2+4\\right)\\boldsymbol{i}+\\left(-3+5\\right)\\boldsymbol{j} \\\\ &amp;=6\\boldsymbol{i}+2\\boldsymbol{j} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Calculating the Component Form of a Vector: Direction<\/h2>\r\nWe have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using [latex]i\\text{and}j[\/latex]. For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.\r\n\r\nCalculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with<strong>\u00a0|<em>v<\/em>|\u00a0<\/strong>replacing<strong><em> r<\/em><\/strong>.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Vector Components in Terms of Magnitude and Direction<\/h3>\r\nGiven a position vector [latex]\\boldsymbol{v}=\\langle x,y\\rangle [\/latex] and a direction angle [latex]\\theta [\/latex],\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\cos \\theta =\\frac{x}{|\\boldsymbol{v}|}&amp;&amp; \\text{and}&amp;&amp; \\sin \\theta =\\frac{y}{|\\boldsymbol{v}|} \\\\ &amp;x=|\\boldsymbol{v}|\\cos \\theta &amp;&amp;&amp;&amp; y=|\\boldsymbol{v}|\\sin \\theta \\end{align}[\/latex]<\/p>\r\nThus, [latex]\\boldsymbol{v}=x\\boldsymbol{i}+y\\boldsymbol{j}=|\\boldsymbol{v}|\\cos \\theta \\boldsymbol{i}+|v|\\sin \\theta \\boldsymbol{j}[\/latex], and magnitude is expressed as [latex]|\\boldsymbol{v}|=\\sqrt{{x}^{2}+{y}^{2}}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 13: Writing a Vector in Terms of Magnitude and Direction<\/h3>\r\nWrite a vector with length 7 at an angle of 135\u00b0 to the positive\u00a0<em>x<\/em>-axis in terms of magnitude and direction.\r\n\r\n[reveal-answer q=\"775938\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"775938\"]\r\n\r\nUsing the conversion formulas [latex]x=|\\boldsymbol{v}|\\cos \\theta i[\/latex] and [latex]y=|\\boldsymbol{v}|\\sin \\theta j[\/latex], we find that\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x&amp;=7\\cos \\left(135^\\circ \\right)\\boldsymbol{i} \\\\ &amp;=-\\frac{7\\sqrt{2}}{2} \\\\ y&amp;=7\\sin \\left(135^\\circ \\right)\\boldsymbol{j} \\\\ &amp;=\\frac{7\\sqrt{2}}{2} \\end{align}[\/latex]<\/p>\r\nThis vector can be written as [latex]v=7\\cos \\left(135^\\circ \\right)i+7\\sin \\left(135^\\circ \\right)j[\/latex] or simplified as\r\n<p style=\"text-align: center;\">[latex]v=-\\dfrac{7\\sqrt{2}}{2}\\boldsymbol{i}+\\dfrac{7\\sqrt{2}}{2}\\boldsymbol{j}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nA vector travels from the origin to the point [latex]\\left(3,5\\right)[\/latex]. Write the vector in terms of magnitude and direction.\r\n\r\n[reveal-answer q=\"365862\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"365862\"]\r\n\r\n[latex]\\boldsymbol{v}=\\sqrt{34}\\cos \\left(59^\\circ \\right)\\boldsymbol{i}+\\sqrt{34}\\sin \\left(59^\\circ \\right)\\boldsymbol{j}[\/latex]\r\nMagnitude = [latex]\\sqrt{34}[\/latex]\r\n[latex]\\theta ={\\tan }^{-1}\\left(\\frac{5}{3}\\right)=59.04^\\circ [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149550[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>\u00a0Finding the Dot Product of Two Vectors<\/h2>\r\nAs we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the <em>dot product<\/em> and the <em>cross product<\/em>. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.\r\n\r\nThe dot product of two vectors involves multiplying two vectors together, and the result is a scalar.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Dot Product<\/h3>\r\nThe <strong>dot product<\/strong> of two vectors [latex]\\boldsymbol{v}=\\langle a,b\\rangle [\/latex] and [latex]\\boldsymbol{u}=\\langle c,d\\rangle [\/latex] is the sum of the product of the horizontal components and the product of the vertical components.\r\n<p style=\"text-align: center;\">[latex]\\boldsymbol{v}\\cdot \\boldsymbol{u}=ac+bd[\/latex]<\/p>\r\nTo find the angle between the two vectors, use the formula below.\r\n<p style=\"text-align: center;\">[latex]\\cos \\theta =\\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\cdot \\dfrac{\\boldsymbol{u}}{|\\boldsymbol{u}|}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 14: Finding the Dot Product of Two Vectors<\/h3>\r\nFind the dot product of [latex]\\boldsymbol{v}=\\langle 5,12\\rangle [\/latex] and [latex]\\boldsymbol{u}=\\langle -3,4\\rangle [\/latex].\r\n\r\n[reveal-answer q=\"734442\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734442\"]\r\n\r\nUsing the formula, we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}\\cdot \\boldsymbol{u}&amp;=\\langle 5,12\\rangle \\cdot \\langle -3,4\\rangle \\\\ &amp;=5\\cdot \\left(-3\\right)+12\\cdot 4 \\\\ &amp;=-15+48 \\\\ &amp;=33 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 15: Finding the Dot Product of Two Vectors and the Angle between Them<\/h3>\r\nFind the dot product of <strong><em>v<\/em><\/strong><sub>1<\/sub> = 5<strong><em>i<\/em><\/strong> + 2<strong><em>j<\/em><\/strong> and <strong><em>v<\/em><\/strong><sub>2<\/sub> = 3<strong><em>i<\/em><\/strong> + 7<strong><em>j<\/em><\/strong>. Then, find the angle between the two vectors.\r\n\r\n[reveal-answer q=\"817790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"817790\"]\r\n\r\nFinding the dot product, we multiply corresponding components.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{1}\\cdot {\\boldsymbol{v}}_{2}&amp;=\\langle 5,2\\rangle \\cdot \\langle 3,7\\rangle \\\\ &amp;=5\\cdot 3+2\\cdot 7 \\\\ &amp;=15+14 \\\\ &amp;=29 \\end{align}[\/latex]<\/p>\r\nTo find the angle between them, we use the formula [latex]\\cos \\theta =\\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\cdot \\dfrac{\\boldsymbol{u}}{|\\boldsymbol{u}|}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\cdot \\frac{\\boldsymbol{u}}{|\\boldsymbol{u}|}&amp;=\\langle \\frac{5}{\\sqrt{29}}+\\frac{2}{\\sqrt{29}}\\rangle \\cdot \\langle \\frac{3}{\\sqrt{58}}+\\frac{7}{\\sqrt{58}}\\rangle \\\\ &amp;=\\frac{5}{\\sqrt{29}}\\cdot \\frac{3}{\\sqrt{58}}+\\frac{2}{\\sqrt{29}}\\cdot \\frac{7}{\\sqrt{58}} \\\\ &amp;=\\frac{15}{\\sqrt{1682}}+\\frac{14}{\\sqrt{1682}}=\\frac{29}{\\sqrt{1682}} \\\\ &amp;=0.707107 \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\theta={\\cos }^{-1}\\left(0.707107\\right)=45^\\circ[\/latex]<\/p>\r\n\r\n<div class=\"mceTemp\"><\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181206\/CNX_Precalc_Figure_08_08_0142.jpg\" alt=\"Plot showing the two position vectors (3,7) and (5,2) and the 45 degree angle between them.\" width=\"487\" height=\"403\" \/> <b>Figure 17<\/b>[\/caption]\r\n\r\n[\/hidden-answer]<b><\/b>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 16: Finding the Angle between Two Vectors<\/h3>\r\nFind the angle between [latex]\\boldsymbol{u}=\\langle -3,4\\rangle [\/latex] and [latex]\\boldsymbol{v}=\\langle 5,12\\rangle [\/latex].\r\n\r\n[reveal-answer q=\"953383\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953383\"]\r\n\r\nUsing the formula, [latex]\\theta ={\\cos }^{-1}\\left(\\dfrac{\\boldsymbol{u}}{|\\boldsymbol{u}|}\\cdot \\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\right)[\/latex],\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\left(\\frac{\\boldsymbol{u}}{|\\boldsymbol{u}|}\\cdot \\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\right)&amp;=\\frac{-3\\boldsymbol{i}+4\\boldsymbol{j}}{5}\\cdot \\frac{5\\boldsymbol{i}+12\\boldsymbol{j}}{13} \\\\ &amp;=\\left(-\\frac{3}{5}\\cdot \\frac{5}{13}\\right)+\\left(\\frac{4}{5}\\cdot \\frac{12}{13}\\right) \\\\ &amp;=-\\frac{15}{65}+\\frac{48}{65} \\\\ &amp;=\\frac{33}{65}\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\theta ={\\cos }^{-1}\\left(\\frac{33}{65}\\right) ={59.5}^{\\circ }[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><img style=\"font-weight: bold; background-color: #f5f5f5; font-size: 0.9em; text-align: left;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181209\/CNX_Precalc_Figure_08_08_0132.jpg\" alt=\"Plot showing the two position vectors (-3,4) and (5,12) and the 59.5 degree angle between them.\" width=\"487\" height=\"628\" \/><\/p>\r\n<b>Figure 18<\/b>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 17: Finding Ground Speed and Bearing Using Vectors<\/h3>\r\nWe now have the tools to solve the problem we introduced in the opening of the section.\r\n\r\nAn airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140\u00b0. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181212\/CNX_Precalc_Figure_08_08_0152.jpg\" alt=\"Image of a plan flying SE at 140 degrees and the north wind blowing.\" width=\"487\" height=\"462\" \/> <b>Figure 19<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"612904\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"612904\"]\r\n\r\nThe ground speed is represented by [latex]x[\/latex] in the diagram, and we need to find the angle [latex]\\alpha [\/latex] in order to calculate the adjusted bearing, which will be [latex]140^\\circ +\\alpha [\/latex].\r\n\r\nNotice in Figure 19, that angle [latex]BCO[\/latex] must be equal to angle [latex]AOC[\/latex] by the rule of alternating interior angles, so angle [latex]BCO[\/latex] is 140\u00b0. We can find [latex]x[\/latex] by the Law of Cosines:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{x}^{2}&amp;={\\left(16.2\\right)}^{2}+{\\left(200\\right)}^{2}-2\\left(16.2\\right)\\left(200\\right)\\cos \\left(140^\\circ \\right) \\\\ &amp;=45,226.41 \\\\ x&amp;=\\sqrt{45,226.41} \\\\ &amp;=212.7 \\end{align}[\/latex]<\/p>\r\nThe ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\sin \\alpha }{16.2}&amp;=\\frac{\\sin \\left(140^\\circ \\right)}{212.7} \\\\ \\sin \\alpha &amp;=\\frac{16.2\\sin \\left(140^\\circ \\right)}{212.7} \\\\ &amp;=0.04896 \\\\ {\\sin }^{-1}&amp;\\left(0.04896\\right)=2.8^\\circ \\end{align}[\/latex]<\/p>\r\nTherefore, the plane has a SE bearing of 140\u00b0+2.8\u00b0=142.8\u00b0. The ground speed is 212.7 miles per hour.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>The position vector has its initial point at the origin.<\/li>\r\n \t<li>If the position vector is the same for two vectors, they are equal.<\/li>\r\n \t<li>Vectors are defined by their magnitude and direction.<\/li>\r\n \t<li>If two vectors have the same magnitude and direction, they are equal.<\/li>\r\n \t<li>Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements.<\/li>\r\n \t<li>Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same.<\/li>\r\n \t<li>Vectors are comprised of two components: the horizontal component along the positive <em>x<\/em>-axis, and the vertical component along the positive <em>y<\/em>-axis.<\/li>\r\n \t<li>The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.<\/li>\r\n \t<li>The magnitude of a vector in the rectangular coordinate system is [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\r\n \t<li>In the rectangular coordinate system, unit vectors may be represented in terms of <em><strong> i <\/strong><\/em> and <em><strong> j <\/strong><\/em> where<em><strong> i <\/strong><\/em>represents the horizontal component and<em><strong> j <\/strong><\/em>represents the vertical component. Then, <strong><em>v<\/em><\/strong> = a<strong><em>i<\/em><\/strong> + b<strong><em>j<\/em><\/strong> is a scalar multiple of<em><strong> v <\/strong><\/em>by real numbers [latex]a\\text{ and }b[\/latex].<\/li>\r\n \t<li>Adding and subtracting vectors in terms of <strong><em>i<\/em><\/strong> and <strong><em>j<\/em><\/strong> consists of adding or subtracting corresponding coefficients of <em><strong>i<\/strong><\/em>\u00a0and corresponding coefficients of <strong><em>j<\/em><\/strong>.<\/li>\r\n \t<li>A vector <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em> is written in terms of magnitude and direction as [latex]\\boldsymbol{v}=|\\boldsymbol{v}|\\cos \\theta \\boldsymbol{i}+|\\boldsymbol{v}|\\sin \\theta \\boldsymbol{j}[\/latex].<\/li>\r\n \t<li>The dot product of two vectors is the product of the<em><strong> i <\/strong><\/em>terms plus the product of the<em><strong> j <\/strong><\/em>terms.<\/li>\r\n \t<li>We can use the dot product to find the angle between two vectors.<\/li>\r\n \t<li>Dot products are useful for many types of physics applications.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165133447852\" class=\"definition\">\r\n \t<dt>dot product<\/dt>\r\n \t<dd id=\"fs-id1165133447857\">given two vectors, the sum of the product of the horizontal components and the product of the vertical components<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133447862\" class=\"definition\">\r\n \t<dt>initial point<\/dt>\r\n \t<dd id=\"fs-id1165135369492\">the origin of a vector<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135369497\" class=\"definition\">\r\n \t<dt>magnitude<\/dt>\r\n \t<dd id=\"fs-id1165135369502\">the length of a vector; may represent a quantity such as speed, and is calculated using the Pythagorean Theorem<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135369506\" class=\"definition\">\r\n \t<dt>resultant<\/dt>\r\n \t<dd id=\"fs-id1165135369512\">a vector that results from addition or subtraction of two vectors, or from scalar multiplication<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135238406\" class=\"definition\">\r\n \t<dt>scalar<\/dt>\r\n \t<dd id=\"fs-id1165135238411\">a quantity associated with magnitude but not direction; a constant<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135238414\" class=\"definition\">\r\n \t<dt>scalar multiplication<\/dt>\r\n \t<dd id=\"fs-id1165135238419\">the product of a constant and each component of a vector<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135238423\" class=\"definition\">\r\n \t<dt>standard position<\/dt>\r\n \t<dd id=\"fs-id1165135369538\">the placement of a vector with the initial point at [latex]\\left(0,0\\right)[\/latex] and the terminal point [latex]\\left(a,b\\right)[\/latex], represented by the change in the <em>x<\/em>-coordinates and the change in the <em>y<\/em>-coordinates of the original vector<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134036770\" class=\"definition\">\r\n \t<dt>terminal point<\/dt>\r\n \t<dd id=\"fs-id1165133243502\">the end point of a vector, usually represented by an arrow indicating its direction<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133243505\" class=\"definition\">\r\n \t<dt>unit vector<\/dt>\r\n \t<dd id=\"fs-id1165133243510\">a vector that begins at the origin and has magnitude of 1; the horizontal unit vector runs along the <em>x<\/em>-axis and is defined as [latex]{v}_{1}=\\langle 1,0\\rangle [\/latex] the vertical unit vector runs along the <em>y<\/em>-axis and is defined as [latex]{v}_{2}=\\langle 0,1\\rangle [\/latex].<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165131906701\" class=\"definition\">\r\n \t<dt>vector<\/dt>\r\n \t<dd id=\"fs-id1165131906706\">a quantity associated with both magnitude and direction, represented as a directed line segment with a starting point (initial point) and an end point (terminal point)<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165131906711\" class=\"definition\">\r\n \t<dt>vector addition<\/dt>\r\n \t<dd id=\"fs-id1165135700056\">the sum of two vectors, found by adding corresponding components<\/dd>\r\n<\/dl>\r\n<div id=\"proctor-extension-installed\"><\/div>","rendered":"<div id=\"proctor-360-confirm\" class=\"proctor-360-confirm proctor-hidden\" style=\"display: none;\">\n<p><img decoding=\"async\" class=\"proctor-360-login-img text-center\" style=\"width: 50px;\" src=\"denied:chrome-extension:\/\/hkegehhbmbongohpgmdadkbkmnfokicn\/img\/icon128.png\" alt=\"logo\" \/><\/p>\n<h3 class=\"text-center\">Are you sure to stop exam tracking?<\/h3>\n<div class=\"proctor-actions\"><button id=\"proctor-360-stop-exam\" class=\"proctor-360-stop-exam btn btn-danger\" type=\"submit\">Yes<\/button> <button id=\"proctor-360-cancel\" class=\"proctor-360-cancel btn btn-success\" type=\"submit\">No<\/button><\/div>\n<\/div>\n<div id=\"proctor-cntnr\"><\/div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>View vectors geometrically.<\/li>\n<li>Find magnitude and direction.<\/li>\n<li>Perform vector addition and scalar multiplication.<\/li>\n<li>Find the component form of a vector.<\/li>\n<li>Find the unit vector in the direction of [latex]\\boldsymbol{v}[\/latex] .<\/li>\n<li>Perform operations with vectors in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex] .<\/li>\n<li>Find the dot product of two vectors.<\/li>\n<\/ul>\n<\/div>\n<p>An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140\u00b0. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 1. What are the ground speed and actual bearing of the plane?<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181127\/CNX_Precalc_Figure_08_08_0012.jpg\" alt=\"Image of a plan flying SE at 140 degrees and the north wind blowing\" width=\"487\" height=\"462\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane\u2019s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let\u2019s examine the basics of vectors.<\/p>\n<h2>A Geometric View of Vectors<\/h2>\n<p>A <strong>vector<\/strong> is a specific quantity drawn as a line segment with an arrowhead at one end. It has an <strong>initial point<\/strong>, where it begins, and a <strong>terminal point<\/strong>, where it ends. A vector is defined by its <strong>magnitude<\/strong>, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:<\/p>\n<ul>\n<li>Lower case, boldfaced type, with or without an arrow on top such as [latex]\\boldsymbol{v,u,w,\\stackrel{\\to }{v},\\stackrel{\\to }{u},\\stackrel{\\to }{w}}[\/latex].<\/li>\n<li>Given initial point [latex]P[\/latex] and terminal point [latex]Q[\/latex], a vector can be represented as [latex]\\stackrel{\\to }{PQ}[\/latex]. The arrowhead on top is what indicates that it is not just a line, but a directed line segment.<\/li>\n<li>Given an initial point of [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(a,b\\right)[\/latex], a vector may be represented as [latex]\\langle a,b\\rangle[\/latex].<\/li>\n<\/ul>\n<p>This last symbol [latex]\\langle a,b\\rangle[\/latex] has special significance. It is called the <strong>standard position<\/strong>. The <strong>position vector<\/strong> has an initial point [latex]\\left(0,0\\right)[\/latex] and a terminal point [latex]\\langle a,b\\rangle[\/latex]. To change any vector into the position vector, we think about the change in the <em>x<\/em>-coordinates and the change in the <em>y<\/em>-coordinates. Thus, if the initial point of a vector [latex]\\stackrel{\\to }{CD}[\/latex] is [latex]C\\left({x}_{1},{y}_{1}\\right)[\/latex] and the terminal point is [latex]D\\left({x}_{2},{y}_{2}\\right)[\/latex], then the position vector is found by calculating<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\stackrel{\\to }{AB}&=\\langle {x}_{2}-{x}_{1},{y}_{2}-{y}_{1}\\rangle \\\\ &=\\langle a,b\\rangle \\end{align}[\/latex]<\/div>\n<p>In Figure 2, we see the original vector [latex]\\stackrel{\\to }{CD}[\/latex] and the position vector [latex]\\stackrel{\\to }{AB}[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181129\/CNX_Precalc_Figure_08_08_0032.jpg\" alt=\"Plot of the original vector CD in blue and the position vector AB in orange extending from the origin.\" width=\"487\" height=\"290\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Properties of Vectors<\/h3>\n<p>A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at [latex]\\left(0,0\\right)[\/latex] and is identified by its terminal point [latex]\\langle a,b\\rangle[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Find the Position Vector<\/h3>\n<p>Consider the vector whose initial point is [latex]P\\left(2,3\\right)[\/latex] and terminal point is [latex]Q\\left(6,4\\right)[\/latex]. Find the position vector.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q454005\">Show Solution<\/span><\/p>\n<div id=\"q454005\" class=\"hidden-answer\" style=\"display: none\">\n<p>The position vector is found by subtracting one <em>x<\/em>-coordinate from the other <em>x<\/em>-coordinate, and one <em>y<\/em>-coordinate from the other <em>y<\/em>-coordinate. Thus<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&=\\langle 6 - 2,4 - 3\\rangle \\\\ &=\\langle 4,1\\rangle \\end{align}[\/latex]<\/p>\n<p>The position vector begins at [latex]\\left(0,0\\right)[\/latex] and terminates at [latex]\\left(4,1\\right)[\/latex]. The graphs of both vectors are shown in Figure 3.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181132\/CNX_Precalc_Figure_08_08_0222.jpg\" alt=\"Plot of the original vector in blue and the position vector in orange extending from the origin.\" width=\"487\" height=\"349\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<p>We see that the position vector is [latex]\\langle 4,1\\rangle[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Drawing a Vector with the Given Criteria and Its Equivalent Position Vector<\/h3>\n<p>Find the position vector given that vector<em><strong> v <\/strong><\/em>has an initial point at [latex]\\left(-3,2\\right)[\/latex] and a terminal point at [latex]\\left(4,5\\right)[\/latex], then graph both vectors in the same plane.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q823850\">Show Solution<\/span><\/p>\n<div id=\"q823850\" class=\"hidden-answer\" style=\"display: none\">\n<p>The position vector is found using the following calculation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&=\\langle 4-\\left(-3\\right),5 - 2\\rangle \\\\ &=\\langle 7,3\\rangle \\end{align}[\/latex]<\/p>\n<p>Thus, the position vector begins at [latex]\\left(0,0\\right)[\/latex] and terminates at [latex]\\left(7,3\\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181134\/CNX_Precalc_Figure_08_08_004n2.jpg\" alt=\"Plot of the two given vectors their same position vector.\" width=\"487\" height=\"328\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Draw a vector [latex]\\boldsymbol{v}[\/latex] that connects from the origin to the point [latex]\\left(3,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q884783\">Show Solution<\/span><\/p>\n<div id=\"q884783\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181235\/CNX_Precalc_Figure_08_08_0062.jpg\" alt=\"A vector from the origin to (3,5) - a line with an arrow at the (3,5) endpoint.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173913\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173913&theme=oea&iframe_resize_id=ohm173913\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Finding Magnitude and Direction<\/h2>\n<p>To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Magnitude and Direction of a Vector<\/h3>\n<p>Given a position vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle[\/latex], the magnitude is found by [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]. The direction is equal to the angle formed with the <em>x<\/em>-axis, or with the <em>y<\/em>-axis, depending on the application. For a position vector, the direction is found by [latex]\\tan \\theta =\\left(\\frac{b}{a}\\right)\\Rightarrow \\theta ={\\tan }^{-1}\\left(\\frac{b}{a}\\right)[\/latex], as illustrated in Figure 5.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181137\/CNX_Precalc_Figure_08_08_017new2.jpg\" alt=\"Standard plot of a position vector (a,b) with magnitude |v| extending into Q1 at theta degrees.\" width=\"487\" height=\"216\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<p>Two vectors <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong> are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Finding the Magnitude and Direction of a Vector<\/h3>\n<p>Find the magnitude and direction of the vector with initial point [latex]P\\left(-8,1\\right)[\/latex] and terminal point [latex]Q\\left(-2,-5\\right)[\/latex]. Draw the vector.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q241600\">Show Solution<\/span><\/p>\n<div id=\"q241600\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find the <strong>position vector<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{u}&=\\langle -2,-\\left(-8\\right),-5 - 1\\rangle \\\\ &=\\langle 6,-6\\rangle \\end{align}[\/latex]<\/p>\n<p>We use the Pythagorean Theorem to find the magnitude.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{u}|&=\\sqrt{{\\left(6\\right)}^{2}+{\\left(-6\\right)}^{2}} \\\\ &=\\sqrt{72} \\\\ &=6\\sqrt{2} \\end{align}[\/latex]<\/p>\n<p>The direction is given as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\tan \\theta =\\frac{-6}{6}=-1\\\\ &\\theta ={\\tan }^{-1}\\left(-1\\right) =-45^\\circ \\end{align}[\/latex]<\/p>\n<p>However, the angle terminates in the fourth quadrant, so we add 360\u00b0 to obtain a positive angle. Thus, [latex]-45^\\circ +360^\\circ =315^\\circ[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181140\/CNX_Precalc_Figure_08_08_0182.jpg\" alt=\"Plot of the position vector extending into Q4 from the origin with the magnitude 6rad2.\" width=\"487\" height=\"316\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Showing That Two Vectors Are Equal<\/h3>\n<p>Show that vector <strong><em>v<\/em><\/strong> with <strong>initial point<\/strong> at [latex]\\left(5,-3\\right)[\/latex] and <strong>terminal point<\/strong> at [latex]\\left(-1,2\\right)[\/latex] is equal to vector <strong><em>u<\/em><\/strong> with initial point at [latex]\\left(-1,-3\\right)[\/latex] and terminal point at [latex]\\left(-7,2\\right)[\/latex]. Draw the position vector on the same grid as <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong>. Next, find the magnitude and direction of each vector.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q46978\">Show Solution<\/span><\/p>\n<div id=\"q46978\" class=\"hidden-answer\" style=\"display: none\">\n<p>As shown in Figure 7, draw the vector [latex]\\boldsymbol{v}[\/latex] starting at initial [latex]\\left(5,-3\\right)[\/latex] and terminal point [latex]\\left(-1,2\\right)[\/latex]. Draw the vector [latex]\\boldsymbol{u}[\/latex] with initial point [latex]\\left(-1,-3\\right)[\/latex] and terminal point [latex]\\left(-7,2\\right)[\/latex]. Find the standard position for each.<\/p>\n<p>Next, find and sketch the position vector for <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong>. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&=\\langle -1 - 5,2-\\left(-3\\right)\\rangle \\\\ &=\\langle -6,5\\rangle \\\\ \\text{ } \\\\ \\boldsymbol{u}&=\\langle -7-\\left(-1\\right),2-\\left(-3\\right)\\rangle \\\\ &=\\langle -6,5\\rangle \\end{align}[\/latex]<\/p>\n<p>Since the position vectors are the same, <strong><em>v<\/em><\/strong> and <strong><em>u<\/em><\/strong> are the same.<\/p>\n<p>An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{v}|&=\\sqrt{{\\left(-1 - 5\\right)}^{2}+{\\left(2-\\left(-3\\right)\\right)}^{2}} \\\\ &=\\sqrt{{\\left(-6\\right)}^{2}+{\\left(5\\right)}^{2}} \\\\ &=\\sqrt{36+25} \\\\ &=\\sqrt{61} \\\\ |\\boldsymbol{u}|&=\\sqrt{{\\left(-7-\\left(-1\\right)\\right)}^{2}+{\\left(2-\\left(-3\\right)\\right)}^{2}} \\\\ &=\\sqrt{{\\left(-6\\right)}^{2}+{\\left(5\\right)}^{2}} \\\\ &=\\sqrt{36+25} \\\\ &=\\sqrt{61} \\end{align}[\/latex]<\/p>\n<p>As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\tan \\theta =-\\frac{5}{6}\\\\ &\\theta ={\\tan }^{-1}\\left(-\\frac{5}{6}\\right) =-39.8^\\circ \\end{align}[\/latex]<\/p>\n<p>However, we can see that the position vector terminates in the second quadrant, so we add [latex]180^\\circ[\/latex]. Thus, the direction is [latex]-39.8^\\circ +180^\\circ =140.2^\\circ[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181143\/CNX_Precalc_Figure_08_08_005n2.jpg\" alt=\"Plot of the two given vectors their same position vector.\" width=\"487\" height=\"440\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Performing Vector Addition and Scalar Multiplication<\/h2>\n<p>Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector [latex]\\boldsymbol{u}[\/latex] [latex]=\\langle x,y\\rangle[\/latex] as an arrow or directed line segment from the origin to the point [latex]\\left(x,y\\right)[\/latex], vectors can be situated anywhere in the plane. The sum of two vectors <strong><em>u<\/em><\/strong> and <strong><em>v<\/em><\/strong>, or <strong>vector addition<\/strong>, produces a third vector <strong><em>u<\/em><\/strong>+ <strong><em>v<\/em><\/strong>, the <strong>resultant<\/strong> vector.<\/p>\n<p>To find <strong><em>u<\/em><\/strong> + <strong><em>v<\/em><\/strong>, we first draw the vector <strong><em>u<\/em><\/strong>, and from the terminal end of <strong><em>u<\/em><\/strong>, we draw the vector <strong><em>v<\/em><\/strong>. In other words, we have the initial point of <strong><em>v<\/em><\/strong> meet the terminal end of <strong><em>u<\/em><\/strong>. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum <strong><em>u<\/em><\/strong> + <strong><em>v<\/em><\/strong> is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of <strong><em>u<\/em><\/strong> to the end of <strong><em>v<\/em><\/strong> in a straight path, as shown in Figure 8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181145\/CNX_Precalc_Figure_08_08_0082.jpg\" alt=\"Diagrams of vector addition and subtraction.\" width=\"487\" height=\"149\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<p>Vector subtraction is similar to vector addition. To find <strong><em>u<\/em><\/strong> \u2212 <strong><em>v<\/em><\/strong>, view it as <strong><em>u<\/em><\/strong> + (\u2212<strong><em>v<\/em><\/strong>). Adding \u2212<strong><em>v<\/em><\/strong> is reversing direction of <strong><em>v<\/em><\/strong> and adding it to the end of <strong><em>u<\/em><\/strong>. The new vector begins at the start of <strong><em>u<\/em><\/strong> and stops at the end point of \u2212<strong><em>v<\/em><\/strong>. See Figure 9\u00a0for a visual that compares vector addition and vector subtraction using <strong>parallelograms<\/strong>.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181147\/CNX_Precalc_Figure_08_08_0092.jpg\" alt=\"Showing vector addition and subtraction with parallelograms. For addition, the base is u, the side is v, the diagonal connecting the start of the base to the end of the side is u+v. For subtraction, thetop is u, the side is -v, and the diagonal connecting the start of the top to the end of the side is u-v.\" width=\"487\" height=\"128\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Adding and Subtracting Vectors<\/h3>\n<p>Given [latex]\\boldsymbol{u}[\/latex] [latex]=\\langle 3,-2\\rangle[\/latex] and [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle -1,4\\rangle[\/latex], find two new vectors <strong><em>u<\/em><\/strong> + <strong><em>v<\/em><\/strong>, and <strong><em>u<\/em><\/strong> \u2212 <strong>v<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q664138\">Show Solution<\/span><\/p>\n<div id=\"q664138\" class=\"hidden-answer\" style=\"display: none\">\n<p>To find the sum of two vectors, we add the components. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{u}+\\boldsymbol{v}&=\\langle 3,-2\\rangle +\\langle -1,4\\rangle \\\\ &=\\langle 3+\\left(-1\\right),-2+4\\rangle \\\\ &=\\langle 2,2\\rangle \\end{align}[\/latex]<\/p>\n<p>See Figure 10(a).<\/p>\n<p>To find the difference of two vectors, add the negative components of [latex]v[\/latex] to [latex]u[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{u}+\\left(-\\boldsymbol{v}\\right)&=\\langle 3,-2\\rangle +\\langle 1,-4\\rangle \\\\ &=\\langle 3+1,-2+\\left(-4\\right)\\rangle \\\\ &=\\langle 4,-6\\rangle \\end{align}[\/latex]<\/p>\n<p>See Figure 10(b).<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181150\/CNX_Precalc_Figure_08_08_0192.jpg\" alt=\"Further diagrams of vector addition and subtraction.\" width=\"731\" height=\"292\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10.<\/b> (a) Sum of two vectors (b) Difference of two vectors<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Multiplying By a Scalar<\/h2>\n<p>While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a <strong>scalar<\/strong>, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Scalar Multiplication<\/h3>\n<p><strong>Scalar multiplication<\/strong> involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle[\/latex] by [latex]k[\/latex] , we have<\/p>\n<p style=\"text-align: center;\">[latex]k\\boldsymbol{v}=\\langle ka,kb\\rangle[\/latex]<\/p>\n<p>Only the magnitude changes, unless [latex]k[\/latex] is negative, and then the vector reverses direction.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Performing Scalar Multiplication<\/h3>\n<p>Given vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle 3,1\\rangle[\/latex], find 3<strong><em>v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] <strong><em>v<\/em>, <\/strong>and \u2212<em><strong>v<\/strong><\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q962766\">Show Solution<\/span><\/p>\n<div id=\"q962766\" class=\"hidden-answer\" style=\"display: none\">\n<p>See Figure 11\u00a0for a geometric interpretation. If [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle 3,1\\rangle[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3\\boldsymbol{v}&=\\langle 3\\cdot 3,3\\cdot 1\\rangle \\\\ &=\\langle 9,3\\rangle \\\\ \\frac{1}{2}\\boldsymbol{v}&=\\langle \\frac{1}{2}\\cdot 3,\\frac{1}{2}\\cdot 1\\rangle \\\\ &=\\langle \\frac{3}{2},\\frac{1}{2}\\rangle \\\\ -\\boldsymbol{v}&=\\langle -3,-1\\rangle \\end{align}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181152\/CNX_Precalc_Figure_08_08_0072.jpg\" alt=\"Showing the effect of scaling a vector: 3x, 1x, .5x, and -1x. The 3x is three times as long, the 1x stays the same, the .5x halves the length, and the -1x reverses the direction of the vector but keeps the length the same. The rest keep the same direction; only the magnitude changes.\" width=\"487\" height=\"367\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice that the vector 3<strong><em>v<\/em><\/strong> is three times the length of <strong><em>v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] [latex]\\boldsymbol{v}[\/latex] is half the length of <strong><em>v<\/em><\/strong>, and \u2013<strong><em>v<\/em><\/strong> is the same length of <strong><em>v<\/em><\/strong>, but in the opposite direction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the <strong>scalar multiple<\/strong>\u00a0[latex]3\\boldsymbol{u}[\/latex] given [latex]\\boldsymbol{u}[\/latex] [latex]=\\langle 5,4\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q453643\">Show Solution<\/span><\/p>\n<div id=\"q453643\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]3\\boldsymbol{u}=\\langle 15,12\\rangle[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Using Vector Addition and Scalar Multiplication to Find a New Vector<\/h3>\n<p>Given [latex]\\boldsymbol{u}=\\langle 3,-2\\rangle[\/latex] and [latex]\\boldsymbol{v}=\\langle -1,4\\rangle[\/latex], find a new vector <strong><em>w<\/em><\/strong> = 3<strong><em>u<\/em><\/strong> + 2<strong>v<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q658063\">Show Solution<\/span><\/p>\n<div id=\"q658063\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we must multiply each vector by the scalar.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3\\boldsymbol{u}&=3\\langle 3,-2\\rangle \\\\ &=\\langle 9,-6\\rangle \\\\ 2\\boldsymbol{v}&=2\\langle -1,4\\rangle \\\\ &=\\langle -2,8\\rangle \\end{align}[\/latex]<\/p>\n<p>Then, add the two together.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{w}&=3\\boldsymbol{u}+2\\boldsymbol{v} \\\\ &=\\langle 9,-6\\rangle +\\langle -2,8\\rangle \\\\ &=\\langle 9 - 2,-6+8\\rangle \\\\ &=\\langle 7,2\\rangle \\end{align}[\/latex]<\/p>\n<p>So, [latex]\\boldsymbol{w}=\\langle 7,2\\rangle[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173921\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173921&theme=oea&iframe_resize_id=ohm173921\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding Component Form<\/h2>\n<p>In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the [latex]x[\/latex] direction, and the vertical component is the [latex]y[\/latex] direction. For example, we can see in the graph in Figure 12\u00a0that the position vector [latex]\\langle 2,3\\rangle[\/latex] comes from adding the vectors <strong><em>v<\/em><\/strong><sub>1<\/sub> and <strong><em>v<\/em><\/strong><sub>2<\/sub>. We have <strong><em>v<\/em><\/strong><sub>1<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(2,0\\right)[\/latex].<\/p>\n<div>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{1}&=\\langle 2 - 0,0 - 0\\rangle \\\\ &=\\langle 2,0\\rangle \\end{align}[\/latex]<\/p>\n<p>We also have <strong><em>v<\/em><\/strong><sub>2<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(0,3\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{2}&=\\langle 0 - 0,3 - 0\\rangle \\\\ &=\\langle 0,3\\rangle \\end{align}[\/latex]<\/div>\n<p>Therefore, the position vector is<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&=\\langle 2+0,3+0\\rangle \\\\ &=\\langle 2,3\\rangle \\end{align}[\/latex]<\/div>\n<p>Using the Pythagorean Theorem, the magnitude of <strong><em>v<\/em><\/strong><sub>1<\/sub> is 2, and the magnitude of <strong><em>v<\/em><\/strong><sub>2<\/sub> is 3. To find the magnitude of <strong><em>v<\/em><\/strong>, use the formula with the position vector.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{v}|&=\\sqrt{|{\\boldsymbol{v}}_{1}{|}^{2}+|{\\boldsymbol{v}}_{2}{|}^{2}} \\\\ &=\\sqrt{{2}^{2}+{3}^{2}} \\\\ &=\\sqrt{13} \\end{align}[\/latex]<\/div>\n<p>The magnitude of <strong><em>v<\/em><\/strong> is [latex]\\sqrt{13}[\/latex]. To find the direction, we use the tangent function [latex]\\tan \\theta =\\frac{y}{x}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&\\tan \\theta =\\frac{{\\boldsymbol{v}}_{2}}{{\\boldsymbol{v}}_{1}} \\\\ &\\tan \\theta =\\frac{3}{2} \\\\ &\\theta ={\\tan }^{-1}\\left(\\frac{3}{2}\\right)=56.3^\\circ \\end{align}[\/latex]<\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181154\/CNX_Precalc_Figure_08_08_0202.jpg\" alt=\"Diagram of a vector in root position with its horizontal and vertical components.\" width=\"487\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12<\/b><\/p>\n<\/div>\n<p>Thus, the magnitude of [latex]\\boldsymbol{v}[\/latex] is [latex]\\sqrt{13}[\/latex] and the direction is [latex]{56.3}^{\\circ }[\/latex] off the horizontal.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 8: Finding the Components of the Vector<\/h3>\n<p>Find the components of the vector [latex]\\boldsymbol{v}[\/latex] with initial point [latex]\\left(3,2\\right)[\/latex] and terminal point [latex]\\left(7,4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q171727\">Show Solution<\/span><\/p>\n<div id=\"q171727\" class=\"hidden-answer\" style=\"display: none\">\n<p>First find the standard position.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&=\\langle 7 - 3,4 - 2\\rangle \\\\ &=\\langle 4,2\\rangle \\end{align}[\/latex]<\/p>\n<p>See the illustration in Figure 13.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181157\/CNX_Precalc_Figure_08_08_0212.jpg\" alt=\"Diagram of a vector in root position with its horizontal (4,0) and vertical (0,2) components.\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 13<\/b><\/p>\n<\/div>\n<p>The horizontal component is [latex]{\\boldsymbol{v}}_{1}=\\langle 4,0\\rangle[\/latex] and the vertical component is [latex]{\\boldsymbol{v}}_{2}=\\langle 0,2\\rangle[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Finding the Unit Vector in the Direction of v<\/h2>\n<p>In addition to finding a vector\u2019s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a <strong>unit vector<\/strong>. We can then preserve the direction of the original vector while simplifying calculations.<\/p>\n<p>Unit vectors are defined in terms of components. The horizontal unit vector is written as [latex]\\boldsymbol{i}=\\langle 1,0\\rangle[\/latex] and is directed along the positive horizontal axis. The vertical unit vector is written as [latex]\\boldsymbol{j}=\\langle 0,1\\rangle[\/latex] and is directed along the positive vertical axis.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181159\/CNX_Precalc_Figure_08_08_011n2.jpg\" alt=\"Plot showing the unit vectors i=91,0) and j=(0,1)\" width=\"487\" height=\"253\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Unit Vectors<\/h3>\n<p>If [latex]\\boldsymbol{v}[\/latex] is a nonzero vector, then [latex]\\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}[\/latex] is a unit vector in the direction of [latex]\\boldsymbol{v}[\/latex]. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Finding the Unit Vector in the Direction of <em>v<\/em><\/h3>\n<p>Find a unit vector in the same direction as [latex]\\boldsymbol{v}=\\langle -5,12\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q326943\">Show Solution<\/span><\/p>\n<div id=\"q326943\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we will find the magnitude.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}|\\boldsymbol{v}|&=\\sqrt{{\\left(-5\\right)}^{2}+{\\left(12\\right)}^{2}} \\\\ &=\\sqrt{25+144} \\\\ &=\\sqrt{169} \\\\ &=13\\end{align}[\/latex]<\/p>\n<p>Then we divide each component by [latex]|v|[\/latex], which gives a unit vector in the same direction as <strong>v<\/strong>:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}=-\\frac{5}{13}i+\\frac{12}{13}j[\/latex]<\/p>\n<p>or, in component form<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}=\\langle -\\frac{5}{13},\\frac{12}{13}\\rangle[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181202\/CNX_Precalc_Figure_08_08_0122.jpg\" alt=\"Plot showing the unit vector (-5\/13, 12\/13) in the direction of (-5, 12)\" width=\"487\" height=\"628\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<p>Verify that the magnitude of the unit vector equals 1. The magnitude of [latex]-\\frac{5}{13}\\boldsymbol{i}+\\frac{12}{13}\\boldsymbol{j}[\/latex] is given as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{{\\left(-\\frac{5}{13}\\right)}^{2}+{\\left(\\frac{12}{13}\\right)}^{2}}&=\\sqrt{\\frac{25}{169}+\\frac{144}{169}} \\\\ &=\\sqrt{\\frac{169}{169}}=1 \\end{align}[\/latex]<\/p>\n<p>The vector\u00a0[latex]u=\\frac{-5}{13}[\/latex] <strong><em>i<\/em><\/strong> [latex]+\\frac{12}{13}[\/latex] <strong><em>j<\/em><\/strong> is the unit vector in the same direction as <strong><em>v<\/em><\/strong> [latex]=\\langle -5,12\\rangle[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Performing Operations with Vectors in Terms of i and j<\/h2>\n<p>So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of <strong><em>i<\/em><\/strong> and <strong>j<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Vectors in the Rectangular Plane<\/h3>\n<p>Given a vector<strong> [latex]v[\/latex] <\/strong>with initial point [latex]P=\\left({x}_{1},{y}_{1}\\right)[\/latex] and terminal point [latex]Q=\\left({x}_{2},{y}_{2}\\right)[\/latex], <strong><em>v<\/em><\/strong> is written as<\/p>\n<p style=\"text-align: center;\">[latex]\\boldsymbol{v}=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{1}-{y}_{2}\\right)\\boldsymbol{j}[\/latex]<\/p>\n<p>The position vector from [latex]\\left(0,0\\right)[\/latex] to [latex]\\left(a,b\\right)[\/latex], where [latex]\\left({x}_{2}-{x}_{1}\\right)=a[\/latex] and [latex]\\left({y}_{2}-{y}_{1}\\right)=b[\/latex], is written as <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em>. This vector sum is called a linear combination of the vectors <strong><em>i<\/em><\/strong> and <strong>j<\/strong>.<\/p>\n<p>The magnitude of <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em> is given as [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181204\/CNX_Precalc_Figure_08_08_010new2.jpg\" alt=\"Plot showing vectors in rectangular coordinates in terms of i and j. The position vector v (in orange) extends from the origin to some point (a,b) in Q1. The horizontal (ai) and vertical (bj) components are shown.\" width=\"487\" height=\"237\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 16<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Writing a Vector in Terms of <em>i<\/em> and <em>j<\/em><\/h3>\n<p>Given a vector [latex]\\boldsymbol{v}[\/latex] with initial point [latex]P=\\left(2,-6\\right)[\/latex] and terminal point [latex]Q=\\left(-6,6\\right)[\/latex], write the vector in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex]<strong>.<\/strong><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q997243\">Show Solution<\/span><\/p>\n<div id=\"q997243\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by writing the general form of the vector. Then replace the coordinates with the given values.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{2}-{y}_{1}\\right)\\boldsymbol{j} \\\\ &=\\left(-6 - 2\\right)\\boldsymbol{i}+\\left(6-\\left(-6\\right)\\right)\\boldsymbol{j} \\\\ &=-8\\boldsymbol{i}+12\\boldsymbol{j} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Writing a Vector in Terms of <em>i<\/em> and <em>j<\/em> Using Initial and Terminal Points<\/h3>\n<p>Given initial point [latex]{P}_{1}=\\left(-1,3\\right)[\/latex] and terminal point [latex]{P}_{2}=\\left(2,7\\right)[\/latex], write the vector [latex]\\boldsymbol{v}[\/latex] in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q301789\">Show Solution<\/span><\/p>\n<div id=\"q301789\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by writing the general form of the vector. Then replace the coordinates with the given values.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}&=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{2}-{y}_{1}\\right)\\boldsymbol{j} \\\\ &=\\left(2-\\left(-1\\right)\\right)\\boldsymbol{i}+\\left(7 - 3\\right)\\boldsymbol{j} \\\\ &=3\\boldsymbol{i}+4\\boldsymbol{j} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the vector [latex]\\boldsymbol{u}[\/latex] with initial point [latex]P=\\left(-1,6\\right)[\/latex] and terminal point [latex]Q=\\left(7,-5\\right)[\/latex] in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q445026\">Show Solution<\/span><\/p>\n<div id=\"q445026\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\boldsymbol{u}=8\\boldsymbol{i} - 11\\boldsymbol{j}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Performing Operations on Vectors in Terms of <em>i<\/em> and <em>j<\/em><\/h2>\n<p>When vectors are written in terms of<em><strong> i <\/strong><\/em>and<strong><em> j<\/em><\/strong>, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Adding and Subtracting Vectors in Rectangular Coordinates<\/h3>\n<p>Given <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em> and <strong><em>u<\/em><\/strong> = <em>c<strong>i<\/strong><\/em> + <em>d<strong>j<\/strong><\/em>, then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\boldsymbol{v}+\\boldsymbol{u}=\\left(a+c\\right)\\boldsymbol{i}+\\left(b+d\\right)\\boldsymbol{j}\\\\ \\boldsymbol{v}-\\boldsymbol{u}=\\left(a-c\\right)\\boldsymbol{i}+\\left(b-d\\right)\\boldsymbol{j}\\end{gathered}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 12: Finding the Sum of the Vectors<\/h3>\n<p>Find the sum of [latex]{\\boldsymbol{v}}_{1}=2\\boldsymbol{i} - 3\\boldsymbol{j}[\/latex] and [latex]{\\boldsymbol{v}}_{2}=4\\boldsymbol{i}+5\\boldsymbol{j}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q139436\">Show Solution<\/span><\/p>\n<div id=\"q139436\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{1}+{\\boldsymbol{v}}_{2}&=\\left(2+4\\right)\\boldsymbol{i}+\\left(-3+5\\right)\\boldsymbol{j} \\\\ &=6\\boldsymbol{i}+2\\boldsymbol{j} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Calculating the Component Form of a Vector: Direction<\/h2>\n<p>We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using [latex]i\\text{and}j[\/latex]. For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.<\/p>\n<p>Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with<strong>\u00a0|<em>v<\/em>|\u00a0<\/strong>replacing<strong><em> r<\/em><\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Vector Components in Terms of Magnitude and Direction<\/h3>\n<p>Given a position vector [latex]\\boldsymbol{v}=\\langle x,y\\rangle[\/latex] and a direction angle [latex]\\theta[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\cos \\theta =\\frac{x}{|\\boldsymbol{v}|}&& \\text{and}&& \\sin \\theta =\\frac{y}{|\\boldsymbol{v}|} \\\\ &x=|\\boldsymbol{v}|\\cos \\theta &&&& y=|\\boldsymbol{v}|\\sin \\theta \\end{align}[\/latex]<\/p>\n<p>Thus, [latex]\\boldsymbol{v}=x\\boldsymbol{i}+y\\boldsymbol{j}=|\\boldsymbol{v}|\\cos \\theta \\boldsymbol{i}+|v|\\sin \\theta \\boldsymbol{j}[\/latex], and magnitude is expressed as [latex]|\\boldsymbol{v}|=\\sqrt{{x}^{2}+{y}^{2}}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 13: Writing a Vector in Terms of Magnitude and Direction<\/h3>\n<p>Write a vector with length 7 at an angle of 135\u00b0 to the positive\u00a0<em>x<\/em>-axis in terms of magnitude and direction.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q775938\">Show Solution<\/span><\/p>\n<div id=\"q775938\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the conversion formulas [latex]x=|\\boldsymbol{v}|\\cos \\theta i[\/latex] and [latex]y=|\\boldsymbol{v}|\\sin \\theta j[\/latex], we find that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x&=7\\cos \\left(135^\\circ \\right)\\boldsymbol{i} \\\\ &=-\\frac{7\\sqrt{2}}{2} \\\\ y&=7\\sin \\left(135^\\circ \\right)\\boldsymbol{j} \\\\ &=\\frac{7\\sqrt{2}}{2} \\end{align}[\/latex]<\/p>\n<p>This vector can be written as [latex]v=7\\cos \\left(135^\\circ \\right)i+7\\sin \\left(135^\\circ \\right)j[\/latex] or simplified as<\/p>\n<p style=\"text-align: center;\">[latex]v=-\\dfrac{7\\sqrt{2}}{2}\\boldsymbol{i}+\\dfrac{7\\sqrt{2}}{2}\\boldsymbol{j}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A vector travels from the origin to the point [latex]\\left(3,5\\right)[\/latex]. Write the vector in terms of magnitude and direction.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q365862\">Show Solution<\/span><\/p>\n<div id=\"q365862\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\boldsymbol{v}=\\sqrt{34}\\cos \\left(59^\\circ \\right)\\boldsymbol{i}+\\sqrt{34}\\sin \\left(59^\\circ \\right)\\boldsymbol{j}[\/latex]<br \/>\nMagnitude = [latex]\\sqrt{34}[\/latex]<br \/>\n[latex]\\theta ={\\tan }^{-1}\\left(\\frac{5}{3}\\right)=59.04^\\circ[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149550\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149550&theme=oea&iframe_resize_id=ohm149550\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>\u00a0Finding the Dot Product of Two Vectors<\/h2>\n<p>As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the <em>dot product<\/em> and the <em>cross product<\/em>. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.<\/p>\n<p>The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Dot Product<\/h3>\n<p>The <strong>dot product<\/strong> of two vectors [latex]\\boldsymbol{v}=\\langle a,b\\rangle[\/latex] and [latex]\\boldsymbol{u}=\\langle c,d\\rangle[\/latex] is the sum of the product of the horizontal components and the product of the vertical components.<\/p>\n<p style=\"text-align: center;\">[latex]\\boldsymbol{v}\\cdot \\boldsymbol{u}=ac+bd[\/latex]<\/p>\n<p>To find the angle between the two vectors, use the formula below.<\/p>\n<p style=\"text-align: center;\">[latex]\\cos \\theta =\\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\cdot \\dfrac{\\boldsymbol{u}}{|\\boldsymbol{u}|}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 14: Finding the Dot Product of Two Vectors<\/h3>\n<p>Find the dot product of [latex]\\boldsymbol{v}=\\langle 5,12\\rangle[\/latex] and [latex]\\boldsymbol{u}=\\langle -3,4\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q734442\">Show Solution<\/span><\/p>\n<div id=\"q734442\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\boldsymbol{v}\\cdot \\boldsymbol{u}&=\\langle 5,12\\rangle \\cdot \\langle -3,4\\rangle \\\\ &=5\\cdot \\left(-3\\right)+12\\cdot 4 \\\\ &=-15+48 \\\\ &=33 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 15: Finding the Dot Product of Two Vectors and the Angle between Them<\/h3>\n<p>Find the dot product of <strong><em>v<\/em><\/strong><sub>1<\/sub> = 5<strong><em>i<\/em><\/strong> + 2<strong><em>j<\/em><\/strong> and <strong><em>v<\/em><\/strong><sub>2<\/sub> = 3<strong><em>i<\/em><\/strong> + 7<strong><em>j<\/em><\/strong>. Then, find the angle between the two vectors.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q817790\">Show Solution<\/span><\/p>\n<div id=\"q817790\" class=\"hidden-answer\" style=\"display: none\">\n<p>Finding the dot product, we multiply corresponding components.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\boldsymbol{v}}_{1}\\cdot {\\boldsymbol{v}}_{2}&=\\langle 5,2\\rangle \\cdot \\langle 3,7\\rangle \\\\ &=5\\cdot 3+2\\cdot 7 \\\\ &=15+14 \\\\ &=29 \\end{align}[\/latex]<\/p>\n<p>To find the angle between them, we use the formula [latex]\\cos \\theta =\\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\cdot \\dfrac{\\boldsymbol{u}}{|\\boldsymbol{u}|}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\cdot \\frac{\\boldsymbol{u}}{|\\boldsymbol{u}|}&=\\langle \\frac{5}{\\sqrt{29}}+\\frac{2}{\\sqrt{29}}\\rangle \\cdot \\langle \\frac{3}{\\sqrt{58}}+\\frac{7}{\\sqrt{58}}\\rangle \\\\ &=\\frac{5}{\\sqrt{29}}\\cdot \\frac{3}{\\sqrt{58}}+\\frac{2}{\\sqrt{29}}\\cdot \\frac{7}{\\sqrt{58}} \\\\ &=\\frac{15}{\\sqrt{1682}}+\\frac{14}{\\sqrt{1682}}=\\frac{29}{\\sqrt{1682}} \\\\ &=0.707107 \\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\theta={\\cos }^{-1}\\left(0.707107\\right)=45^\\circ[\/latex]<\/p>\n<div class=\"mceTemp\"><\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181206\/CNX_Precalc_Figure_08_08_0142.jpg\" alt=\"Plot showing the two position vectors (3,7) and (5,2) and the 45 degree angle between them.\" width=\"487\" height=\"403\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 17<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 16: Finding the Angle between Two Vectors<\/h3>\n<p>Find the angle between [latex]\\boldsymbol{u}=\\langle -3,4\\rangle[\/latex] and [latex]\\boldsymbol{v}=\\langle 5,12\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953383\">Show Solution<\/span><\/p>\n<div id=\"q953383\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, [latex]\\theta ={\\cos }^{-1}\\left(\\dfrac{\\boldsymbol{u}}{|\\boldsymbol{u}|}\\cdot \\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\right)[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\left(\\frac{\\boldsymbol{u}}{|\\boldsymbol{u}|}\\cdot \\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}\\right)&=\\frac{-3\\boldsymbol{i}+4\\boldsymbol{j}}{5}\\cdot \\frac{5\\boldsymbol{i}+12\\boldsymbol{j}}{13} \\\\ &=\\left(-\\frac{3}{5}\\cdot \\frac{5}{13}\\right)+\\left(\\frac{4}{5}\\cdot \\frac{12}{13}\\right) \\\\ &=-\\frac{15}{65}+\\frac{48}{65} \\\\ &=\\frac{33}{65}\\end{align}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\theta ={\\cos }^{-1}\\left(\\frac{33}{65}\\right) ={59.5}^{\\circ }[\/latex]<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" style=\"font-weight: bold; background-color: #f5f5f5; font-size: 0.9em; text-align: left;\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181209\/CNX_Precalc_Figure_08_08_0132.jpg\" alt=\"Plot showing the two position vectors (-3,4) and (5,12) and the 59.5 degree angle between them.\" width=\"487\" height=\"628\" \/><\/p>\n<p><b>Figure 18<\/b><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 17: Finding Ground Speed and Bearing Using Vectors<\/h3>\n<p>We now have the tools to solve the problem we introduced in the opening of the section.<\/p>\n<p>An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140\u00b0. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane?<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27181212\/CNX_Precalc_Figure_08_08_0152.jpg\" alt=\"Image of a plan flying SE at 140 degrees and the north wind blowing.\" width=\"487\" height=\"462\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 19<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q612904\">Show Solution<\/span><\/p>\n<div id=\"q612904\" class=\"hidden-answer\" style=\"display: none\">\n<p>The ground speed is represented by [latex]x[\/latex] in the diagram, and we need to find the angle [latex]\\alpha[\/latex] in order to calculate the adjusted bearing, which will be [latex]140^\\circ +\\alpha[\/latex].<\/p>\n<p>Notice in Figure 19, that angle [latex]BCO[\/latex] must be equal to angle [latex]AOC[\/latex] by the rule of alternating interior angles, so angle [latex]BCO[\/latex] is 140\u00b0. We can find [latex]x[\/latex] by the Law of Cosines:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{x}^{2}&={\\left(16.2\\right)}^{2}+{\\left(200\\right)}^{2}-2\\left(16.2\\right)\\left(200\\right)\\cos \\left(140^\\circ \\right) \\\\ &=45,226.41 \\\\ x&=\\sqrt{45,226.41} \\\\ &=212.7 \\end{align}[\/latex]<\/p>\n<p>The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{\\sin \\alpha }{16.2}&=\\frac{\\sin \\left(140^\\circ \\right)}{212.7} \\\\ \\sin \\alpha &=\\frac{16.2\\sin \\left(140^\\circ \\right)}{212.7} \\\\ &=0.04896 \\\\ {\\sin }^{-1}&\\left(0.04896\\right)=2.8^\\circ \\end{align}[\/latex]<\/p>\n<p>Therefore, the plane has a SE bearing of 140\u00b0+2.8\u00b0=142.8\u00b0. The ground speed is 212.7 miles per hour.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The position vector has its initial point at the origin.<\/li>\n<li>If the position vector is the same for two vectors, they are equal.<\/li>\n<li>Vectors are defined by their magnitude and direction.<\/li>\n<li>If two vectors have the same magnitude and direction, they are equal.<\/li>\n<li>Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements.<\/li>\n<li>Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same.<\/li>\n<li>Vectors are comprised of two components: the horizontal component along the positive <em>x<\/em>-axis, and the vertical component along the positive <em>y<\/em>-axis.<\/li>\n<li>The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.<\/li>\n<li>The magnitude of a vector in the rectangular coordinate system is [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].<\/li>\n<li>In the rectangular coordinate system, unit vectors may be represented in terms of <em><strong> i <\/strong><\/em> and <em><strong> j <\/strong><\/em> where<em><strong> i <\/strong><\/em>represents the horizontal component and<em><strong> j <\/strong><\/em>represents the vertical component. Then, <strong><em>v<\/em><\/strong> = a<strong><em>i<\/em><\/strong> + b<strong><em>j<\/em><\/strong> is a scalar multiple of<em><strong> v <\/strong><\/em>by real numbers [latex]a\\text{ and }b[\/latex].<\/li>\n<li>Adding and subtracting vectors in terms of <strong><em>i<\/em><\/strong> and <strong><em>j<\/em><\/strong> consists of adding or subtracting corresponding coefficients of <em><strong>i<\/strong><\/em>\u00a0and corresponding coefficients of <strong><em>j<\/em><\/strong>.<\/li>\n<li>A vector <strong><em>v<\/em><\/strong> = <em>a<strong>i<\/strong><\/em> + <em>b<strong>j<\/strong><\/em> is written in terms of magnitude and direction as [latex]\\boldsymbol{v}=|\\boldsymbol{v}|\\cos \\theta \\boldsymbol{i}+|\\boldsymbol{v}|\\sin \\theta \\boldsymbol{j}[\/latex].<\/li>\n<li>The dot product of two vectors is the product of the<em><strong> i <\/strong><\/em>terms plus the product of the<em><strong> j <\/strong><\/em>terms.<\/li>\n<li>We can use the dot product to find the angle between two vectors.<\/li>\n<li>Dot products are useful for many types of physics applications.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165133447852\" class=\"definition\">\n<dt>dot product<\/dt>\n<dd id=\"fs-id1165133447857\">given two vectors, the sum of the product of the horizontal components and the product of the vertical components<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133447862\" class=\"definition\">\n<dt>initial point<\/dt>\n<dd id=\"fs-id1165135369492\">the origin of a vector<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135369497\" class=\"definition\">\n<dt>magnitude<\/dt>\n<dd id=\"fs-id1165135369502\">the length of a vector; may represent a quantity such as speed, and is calculated using the Pythagorean Theorem<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135369506\" class=\"definition\">\n<dt>resultant<\/dt>\n<dd id=\"fs-id1165135369512\">a vector that results from addition or subtraction of two vectors, or from scalar multiplication<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135238406\" class=\"definition\">\n<dt>scalar<\/dt>\n<dd id=\"fs-id1165135238411\">a quantity associated with magnitude but not direction; a constant<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135238414\" class=\"definition\">\n<dt>scalar multiplication<\/dt>\n<dd id=\"fs-id1165135238419\">the product of a constant and each component of a vector<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135238423\" class=\"definition\">\n<dt>standard position<\/dt>\n<dd id=\"fs-id1165135369538\">the placement of a vector with the initial point at [latex]\\left(0,0\\right)[\/latex] and the terminal point [latex]\\left(a,b\\right)[\/latex], represented by the change in the <em>x<\/em>-coordinates and the change in the <em>y<\/em>-coordinates of the original vector<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134036770\" class=\"definition\">\n<dt>terminal point<\/dt>\n<dd id=\"fs-id1165133243502\">the end point of a vector, usually represented by an arrow indicating its direction<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133243505\" class=\"definition\">\n<dt>unit vector<\/dt>\n<dd id=\"fs-id1165133243510\">a vector that begins at the origin and has magnitude of 1; the horizontal unit vector runs along the <em>x<\/em>-axis and is defined as [latex]{v}_{1}=\\langle 1,0\\rangle[\/latex] the vertical unit vector runs along the <em>y<\/em>-axis and is defined as [latex]{v}_{2}=\\langle 0,1\\rangle[\/latex].<\/dd>\n<\/dl>\n<dl id=\"fs-id1165131906701\" class=\"definition\">\n<dt>vector<\/dt>\n<dd id=\"fs-id1165131906706\">a quantity associated with both magnitude and direction, represented as a directed line segment with a starting point (initial point) and an end point (terminal point)<\/dd>\n<\/dl>\n<dl id=\"fs-id1165131906711\" class=\"definition\">\n<dt>vector addition<\/dt>\n<dd id=\"fs-id1165135700056\">the sum of two vectors, found by adding corresponding components<\/dd>\n<\/dl>\n<div id=\"proctor-extension-installed\"><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1954\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":708740,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1954","chapter","type-chapter","status-publish","hentry"],"part":1951,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1954","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/users\/708740"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1954\/revisions"}],"predecessor-version":[{"id":2466,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1954\/revisions\/2466"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/parts\/1951"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1954\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/media?parent=1954"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1954"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/contributor?post=1954"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/license?post=1954"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}